Question

A golf ball thrown at an angle of 20 relative to the ground at a speed of 40 m/s. if the range of the motion is 55m. what is the maximum height that the ball reaches.

Answer 1

Maximum height of the projectile is 5.00 m

Step-by-step explanation:

As we know that the projectile range is the net horizontal distance moved by the object

Here we know that angle of projection is

`\theta = 20^o`

velocity of projection is 40 m/s

now we know the formula of range as

`R = (v^2 sin2\theta)/(g)`

so we have

`55 = (40^2 sin(2* 20))/(g)`

now we have

`g = 18.7 m/s^2`

Now maximum height of the projectile is given as

`H = (v^2 sin^2\theta)/(2g)`

`H = (40^2sin^220)/(2(18.7))`

`H = 5 m`